document.write( "Question 5053: Q. Prove the following identity: 1 divided by 1+cosx = 1/sin^2x - 1/sinxtanx
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document.write( "I am at the dead end but these are my steps:
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document.write( "R.S.
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document.write( "1 divided by(sinx)(sinx)-1 divided by (sinx)(sinx/cosx)=
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document.write( "sin^2x+cos^2x divided by sin^2x - sin^2x+cos^2xdivided by (sin^2x/cosx)=
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document.write( "(sinx)(sinx)+(cosx)(cos)divided by (sinx)(sinx) - (sinx)(sinx) + (cosx)(cosx)(cosx)divided by (sinx)(sinx)=
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document.write( "sin^2x + cos^2x - sin^2x + cos^2x(cosx) divided by (sinx)(sinx)=
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document.write( "and here either
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document.write( "1 - cosx divided by sin^2x
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document.write( "or
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document.write( "2cos^2x(cosx)divided by sin^2x
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document.write( "--------
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document.write( "thanks mmm
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document.write( "
Algebra.Com's Answer #2520 by longjonsilver(2297)![]() ![]() You can put this solution on YOUR website! in any proof, pick one of the 2 sides and manipulate it until you get the other side. Always start with the more complicated side too!\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Looking at your proof: \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "jon. \n" ); document.write( " \n" ); document.write( " |